Integrand size = 11, antiderivative size = 41 \[ \int \frac {x^3}{1+x^3} \, dx=x+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {327, 206, 31, 648, 632, 210, 642} \[ \int \frac {x^3}{1+x^3} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} \log \left (x^2-x+1\right )+x-\frac {1}{3} \log (x+1) \]
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Rule 31
Rule 206
Rule 210
Rule 327
Rule 632
Rule 642
Rule 648
Rubi steps \begin{align*} \text {integral}& = x-\int \frac {1}{1+x^3} \, dx \\ & = x-\frac {1}{3} \int \frac {1}{1+x} \, dx-\frac {1}{3} \int \frac {2-x}{1-x+x^2} \, dx \\ & = x-\frac {1}{3} \log (1+x)+\frac {1}{6} \int \frac {-1+2 x}{1-x+x^2} \, dx-\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx \\ & = x-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right )+\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right ) \\ & = x-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{1+x^3} \, dx=x-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \]
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Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83
method | result | size |
risch | \(x -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\) | \(34\) |
default | \(x +\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+x \right )}{3}\) | \(36\) |
meijerg | \(x -\frac {x \left (\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}-\frac {\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}\) | \(74\) |
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{1+x^3} \, dx=x - \frac {\log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \frac {x^3}{1+x^3} \, dx=x-\frac {\ln \left (x+1\right )}{3}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]
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